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Hi I'm struggling with this exercise

Exercise 3.4.10. Suppose that I and J are two sets, and for all α ∈ I ∪ J let Aα be a set. If I and J are non-empty, show that (⋂α∈I Aα) ∩ (⋂α∈J Aα) = ⋂α∈I∪J Aα.

I know how to show that two sets are equal.

first let y be a arbitrary element.

if y∈ (⋂α∈I Aα) ∩ (⋂α∈J Aα)

then (y∈Aa for all a∈i) and (y∈Aa for all a∈j)

and then

I have to show that (y∈Aa for all a∈i U j). but this stage I can't deduce this statement from the above

statement using rule of inference that I know.

I can explain why this is valid inference in words, but I can't explain by predicate calculus

please someone help me!

ju so
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1 Answers1

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We have, by definition of arbitrary intersection and union of two sets:$$\begin{align} y \in \bigcap_{\alpha \in I\cup J} A_\alpha & \iff \forall \alpha \in I\cup J (y \in A_\alpha)\\& \iff \forall \alpha (\alpha \in I\cup J \rightarrow y \in A_\alpha) \\ & \iff \forall \alpha(\alpha \in I \lor \alpha \in J \rightarrow y\in A_\alpha) \end{align}$$

You have shown $\forall \alpha(\alpha \in I \rightarrow y \in A_\alpha)$ and $\forall \alpha(\alpha \in U \rightarrow y \in A_\alpha)$.

We can prove this using the rule of inference called modus ponens, and by using proof by cases. We want to prove $\forall \alpha(\alpha \in I \lor \alpha \in J \rightarrow y\in A_\alpha)$, so let $\alpha$ be arbitrary. First suppose $\alpha \in I$. Apply modus ponens to this and the statement $\alpha \in I \rightarrow y \in A_\alpha$ (which was already proved) to conclude $y \in A_\alpha$ . Now suppose $\alpha \in J$ and apply modus ponens again, this time using $\alpha \in J \rightarrow y \in A_\alpha$. So $y \in A_\alpha$ holds for an exhaustive list of cases, and we have proven $\forall \alpha(\alpha \in I \lor \alpha \in J \rightarrow y\in A_\alpha)$.

masiewpao
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